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Moving boundary diffusion equation (transformation of coordinates)

  1. Jan 19, 2014 #1
    I'm trying to implement a numerical code for the diffusion equation with moving boundaries. I have no problems with the numerical implementation, but with the transformation of coordinates. In spherical coordinates, the diffusion equation is
    [itex]\frac{\partial c}{\partial t} = D \left(\frac{2}{r}\frac{\partial c}{\partial r} + \frac{\partial^2 c}{\partial r^2} \right)[/itex].

    After scaling these equations with [itex] \phi = c/C_0, \tau = Dt/R_0^2, x = (r-R_0)/R_0 [/itex] I get

    [itex] \frac{\partial\phi}{\partial \tau} = \frac{2}{1+x} \frac{\partial\phi}{\partial x} + \frac{\partial^2 \phi}{\partial x^2} [/itex]

    Now transform the [itex]x[/itex] coordinate to the Landau (not that Landau) coordinate [itex]\eta(x,\tau)[/itex] defined as

    [itex]\eta(x,\tau) = \frac{x - X(\tau)}{X_\infty - X(\tau)}[/itex].

    The idea of the transformation is to move the boundary [itex]X(\tau)[/itex], which is variable only in time [itex]\tau[/itex] and keep the length scale [itex]X_\infty[/itex] fixed. This is particularly useful, for instance, in problems involving a gas bubble losing volume by diffusion, where the bubble's radius is variable in time.

    There's a paper out there (Fischer, Zinovik and Poulikakos 2009) which suggests that the final result should be

    [itex]\frac{\partial \phi}{\partial \tau} = \left[ \frac{2}{(X_\infty - X(\tau))(1+\eta (X_\infty - X(\tau))+ X)} + \frac{1-\eta}{X_\infty - X(\tau)}\frac{dX}{d\tau} \right] \frac{\partial\phi}{\partial\eta} + \frac{1}{(X_\infty - X(\tau))^2} \frac{\partial^2 \phi}{\partial \eta^2}.
    [/itex]

    However, I have some trouble reproducing this result. Can someone guide me through the change of coordinates? I have problems particularly with the second partial derivative.
     
  2. jcsd
  3. Jan 19, 2014 #2

    pasmith

    User Avatar
    Homework Helper

    It helps, to avoid confusion, to change to two new independent variables, so let's take [itex](x,\tau) \to (\eta, \sigma)[/itex] with [itex]\sigma(x,\tau) = \tau[/itex].

    The chain rule gives
    [tex]
    \frac{\partial \phi}{\partial x} = \frac{\partial \phi}{\partial \eta} \frac{\partial \eta}{\partial x} + \frac{\partial \phi}{\partial \sigma} \frac{\partial \sigma}{\partial x}
    = \frac{1}{X_\infty - X(\sigma)} \frac{\partial \phi}{\partial \eta} \\
    \frac{\partial \phi}{\partial \tau} = \frac{\partial \phi}{\partial \eta} \frac{\partial \eta}{\partial \tau} + \frac{\partial \phi}{\partial \sigma} \frac{\partial \sigma}{\partial \tau}
    = -\frac{(1- \eta) X'(\sigma)}{X_\infty - X(\sigma)} \frac{\partial \phi}{\partial \eta} + \frac{\partial \phi}{\partial \sigma}
    [/tex]
    since
    [tex]
    \frac{\partial \eta}{\partial \tau} = \frac{\partial}{\partial \tau} \left(\frac{x - X(\tau)}{X_\infty - X(\tau)}\right) =
    \frac{1}{(X_\infty - X)^2} ((X_\infty - X)(-X') - (x - X)(-X')) \\
    = \frac{-X'(\sigma)}{X_\infty - X(\sigma)} \frac{X_\infty - x}{X_\infty - X(\sigma)}
    = \frac{-X'(\sigma)}{X_\infty - X(\sigma)}(1-\eta)
    [/tex]

    For the second derivative,
    [tex]
    \frac{\partial^2 \phi}{\partial x^2} = \frac{\partial}{\partial x} \frac{\partial \phi}{\partial x}
    = \frac{\partial}{\partial x} \left( \frac{1}{X_\infty - X(\sigma)} \frac{\partial \phi}{\partial \eta} \right) \\
    = \frac{\partial \eta}{\partial x} \frac{\partial}{\partial \eta} \left( \frac{1}{X_\infty - X(\sigma)} \frac{\partial \phi}{\partial \eta} \right)
    + \frac{\partial \sigma}{\partial x} \frac{\partial}{\partial \sigma} \left( \frac{1}{X_\infty - X(\sigma)} \frac{\partial \phi}{\partial \eta} \right) \\
    = \frac{1}{(X_\infty - X(\sigma))^2} \frac{\partial^2 \phi}{\partial \eta^2}
    [/tex]

    Substitution into the original equation and rearrangement will yield the given result.
     
    Last edited: Jan 19, 2014
  4. Jan 19, 2014 #3
    Hi Pasmith, thanks, I understand it now!
     
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